Points A and B are 150 m apart on one bank of a river. Point C is on the other bank of the river. The line of sight distance from A to C forms an angle and 28 degrees with the AB side of the river bank and the line of sight distance form B to C forms an angle of 56 degrees with the AB side of the river bank. How wide is the river?
You have ASA. Solve the triangle. (getting the two other sides, law of sines).
Now, area of the triangle= sqrt(s(s-a)(s-b)(s-c)) where s is the half-perimeter (this is known as Heron of Alexandria's formula).
Now set it equal to 1/2 Base*height,or
area= 1/2 (150)width river
setting the two areas equal, you can find the width of the river
To find the width of the river, we will use trigonometry and the given angles.
Let's consider a right-angled triangle where AB is the base of the triangle, with points A and B forming the right angle. Point C is located on the opposite bank of the river. We are given two angles: 28 degrees at point A and 56 degrees at point B.
First, let's label the width of the river as "x".
We can use the tangent function to find the length of AC:
tan(28°) = AC / x
AC = x * tan(28°)
Similarly, we can use the tangent function to find the length of BC:
tan(56°) = BC / x
BC = x * tan(56°)
The total distance from A to B is given as 150 m. Thus, we can express the width of the river using the equation:
AB = AC + BC + x
150 = x * tan(28°) + x * tan(56°) + x
Now we can solve for x:
150 = x * (tan(28°) + tan(56°) + 1)
By rearranging the equation, we find:
x = 150 / (tan(28°) + tan(56°) + 1)
Using a scientific calculator or trigonometric tables, we can calculate the values of tan(28°) and tan(56°) and then substitute them into the equation to find the width of the river (x).